Inference for categorical data
Renida Kasa
Getting Started
Load packages
In this lab, we will explore and visualize the data using the tidyverse suite of packages, and perform statistical inference using infer. The data can be found in the companion package for OpenIntro resources, openintro.
Let’s load the packages.
library(tidyverse)
library(openintro)
library(infer)The data
You will be analyzing the same dataset as in the previous lab, where you delved into a sample from the Youth Risk Behavior Surveillance System (YRBSS) survey, which uses data from high schoolers to help discover health patterns. The dataset is called yrbss.
- What are the counts within each category for the amount of days these students have texted while driving within the past 30 days?
glimpse(yrbss)## Rows: 13,583
## Columns: 13
## $ age <int> 14, 14, 15, 15, 15, 15, 15, 14, 15, 15, 15, 1…
## $ gender <chr> "female", "female", "female", "female", "fema…
## $ grade <chr> "9", "9", "9", "9", "9", "9", "9", "9", "9", …
## $ hispanic <chr> "not", "not", "hispanic", "not", "not", "not"…
## $ race <chr> "Black or African American", "Black or Africa…
## $ height <dbl> NA, NA, 1.73, 1.60, 1.50, 1.57, 1.65, 1.88, 1…
## $ weight <dbl> NA, NA, 84.37, 55.79, 46.72, 67.13, 131.54, 7…
## $ helmet_12m <chr> "never", "never", "never", "never", "did not …
## $ text_while_driving_30d <chr> "0", NA, "30", "0", "did not drive", "did not…
## $ physically_active_7d <int> 4, 2, 7, 0, 2, 1, 4, 4, 5, 0, 0, 0, 4, 7, 7, …
## $ hours_tv_per_school_day <chr> "5+", "5+", "5+", "2", "3", "5+", "5+", "5+",…
## $ strength_training_7d <int> 0, 0, 0, 0, 1, 0, 2, 0, 3, 0, 3, 0, 0, 7, 7, …
## $ school_night_hours_sleep <chr> "8", "6", "<5", "6", "9", "8", "9", "6", "<5"…yrbss %>%
count(text_while_driving_30d)## # A tibble: 9 × 2
## text_while_driving_30d n
## <chr> <int>
## 1 0 4792
## 2 1-2 925
## 3 10-19 373
## 4 20-29 298
## 5 3-5 493
## 6 30 827
## 7 6-9 311
## 8 did not drive 4646
## 9 <NA> 918In the past 30 days, 4792 students reported that they texted 0 days while driving, 925 reported texting 1-2 days, 493 reported texting 3-5 days, 311 reported texting 6-9 days, 373 reported texting 10-19 days, 298 reported texting 20-29 days, 827 reported texting 30 days, and 4646 reported that they did not drive. There is missing data for 918 students.
- What is the proportion of people who have texted while driving every day in the past 30 days and never wear helmets?
About 6.64% of the respondents who have never worn a helmet have texted while driving every day in the past 30 days.
Remember that you can use filter to limit the dataset to just non-helmet wearers. Here, we will name the dataset no_helmet.
# non helmet wearers:
data('yrbss', package='openintro')
no_helmet <- yrbss %>%
filter(helmet_12m == "never")Also, it may be easier to calculate the proportion if you create a new variable that specifies whether the individual has texted every day while driving over the past 30 days or not. We will call this variable text_ind.
#non helmet + specifified
no_helmet <- no_helmet %>%
mutate(text_ind = ifelse(text_while_driving_30d == "30", "yes", "no"))no_helmet %>%
count(text_ind) %>%
mutate(p = n / sum(n))## # A tibble: 3 × 3
## text_ind n p
## <chr> <int> <dbl>
## 1 no 6040 0.866
## 2 yes 463 0.0664
## 3 <NA> 474 0.0679Inference on proportions
When summarizing the YRBSS, the Centers for Disease Control and Prevention seeks insight into the population parameters. To do this, you can answer the question, “What proportion of people in your sample reported that they have texted while driving each day for the past 30 days?” with a statistic; while the question “What proportion of people on earth have texted while driving each day for the past 30 days?” is answered with an estimate of the parameter.
The inferential tools for estimating population proportion are analogous to those used for means in the last chapter: the confidence interval and the hypothesis test.
no_helmet %>%
drop_na(text_ind) %>% # Drop missing values
specify(response = text_ind, success = "yes") %>%
generate(reps = 1000, type = "bootstrap") %>%
calculate(stat = "prop") %>%
get_ci(level = 0.95)## # A tibble: 1 × 2
## lower_ci upper_ci
## <dbl> <dbl>
## 1 0.0654 0.0775Note that since the goal is to construct an interval estimate for a proportion, it’s necessary to both include the success argument within specify, which accounts for the proportion of non-helmet wearers than have consistently texted while driving the past 30 days, in this example, and that stat within calculate is here “prop”, signaling that you are trying to do some sort of inference on a proportion.
- What is the margin of error for the estimate of the proportion of non-helmet wearers that have texted while driving each day for the past 30 days based on this survey?
margin_of_error <- (0.07719514 - 0.06504306) / 2 #confidence interval/2
round(margin_of_error,3)## [1] 0.006The margin of error is about .006. It can be found by calculating half of the width of the confidence interval.
- Using the
inferpackage, calculate confidence intervals for two other categorical variables (you’ll need to decide which level to call “success”, and report the associated margins of error. Interpret the interval in context of the data. It may be helpful to create new data sets for each of the two countries first, and then use these data sets to construct the confidence intervals.
no_helmet <- no_helmet %>%
mutate(tv_ind = ifelse(hours_tv_per_school_day <= 5, "yes", "no"))
no_helmet %>%
drop_na(tv_ind) %>%
specify(response = tv_ind, success = "yes") %>% #success if student watches <=5 hrs tv per day
generate(reps = 1000, type = "bootstrap") %>%
calculate(stat = "prop") %>%
get_ci(level = 0.95)## # A tibble: 1 × 2
## lower_ci upper_ci
## <dbl> <dbl>
## 1 0.749 0.769width_tv<-0.7690949-0.7483444
print(width_tv)## [1] 0.0207505margin_of_error_tv <- width_tv / 2 #confidence interval/2
round(margin_of_error_tv,2)## [1] 0.01The confidence interval for students who watch less than or equal to 5 hours of television daily is about .75 - .77. The width is about .02, and the margin of error is .01. This means that between 75-77% of the students watch television for less than or equal to 5 hours per day, with a margin of error of 1%, or 76±1% of students.
no_helmet <- no_helmet %>%
mutate(sleep_ind = ifelse(school_night_hours_sleep >= 8, "yes", "no"))
no_helmet %>%
drop_na(sleep_ind) %>%
specify(response = sleep_ind, success = "yes") %>% #success if student sleeps >8 hrs per day
generate(reps = 1000, type = "bootstrap") %>%
calculate(stat = "prop") %>%
get_ci(level = 0.95)## # A tibble: 1 × 2
## lower_ci upper_ci
## <dbl> <dbl>
## 1 0.268 0.289width_sleep<-0.2902354-0.2678117
print(width_sleep)## [1] 0.0224237margin_of_error_sleep <- width_sleep / 2 #confidence interval/2
round(margin_of_error_sleep,2)## [1] 0.01The confidence interval for students who sleep for at least 8 hours or more per day is about .27 - .29. The width is about .02, and the margin of error is .01. This means that between 27-29% of the students sleep for at least 8 hours per night, with a margin of error of 1%, or 28±1% of students.
How does the proportion affect the margin of error?
Imagine you’ve set out to survey 1000 people on two questions: are you at least 6-feet tall? and are you left-handed? Since both of these sample proportions were calculated from the same sample size, they should have the same margin of error, right? Wrong! While the margin of error does change with sample size, it is also affected by the proportion.
Think back to the formula for the standard error: \(SE = \sqrt{p(1-p)/n}\). This is then used in the formula for the margin of error for a 95% confidence interval:
\[ ME = 1.96\times SE = 1.96\times\sqrt{p(1-p)/n} \,. \] Since the population proportion \(p\) is in this \(ME\) formula, it should make sense that the margin of error is in some way dependent on the population proportion. We can visualize this relationship by creating a plot of \(ME\) vs. \(p\).
Since sample size is irrelevant to this discussion, let’s just set it to some value (\(n = 1000\)) and use this value in the following calculations:
n <- 1000The first step is to make a variable p that is a sequence from 0 to 1 with each number incremented by 0.01. You can then create a variable of the margin of error (me) associated with each of these values of p using the familiar approximate formula (\(ME = 2 \times SE\)).
p <- seq(from = 0, to = 1, by = 0.01)
me <- 2 * sqrt(p * (1 - p)/n)Lastly, you can plot the two variables against each other to reveal their relationship. To do so, we need to first put these variables in a data frame that you can call in the ggplot function.
dd <- data.frame(p = p, me = me)
ggplot(data = dd, aes(x = p, y = me)) +
geom_line() +
labs(x = "Population Proportion", y = "Margin of Error")
- Describe the relationship between
pandme. Include the margin of error vs. population proportion plot you constructed in your answer. For a given sample size, for which value ofpis margin of error maximized?
I would describe the shape of this graph as parabolic. As the population proportion increases, the margin of error also increases until a peak is reached. This peak is at 50% of the population, where the margin of error is maximized. Then, as the population increases to 100%, the margin of error decreases.
Success-failure condition
We have emphasized that you must always check conditions before making inference. For inference on proportions, the sample proportion can be assumed to be nearly normal if it is based upon a random sample of independent observations and if both \(np \geq 10\) and \(n(1 - p) \geq 10\). This rule of thumb is easy enough to follow, but it makes you wonder: what’s so special about the number 10?
The short answer is: nothing. You could argue that you would be fine with 9 or that you really should be using 11. What is the “best” value for such a rule of thumb is, at least to some degree, arbitrary. However, when \(np\) and \(n(1-p)\) reaches 10 the sampling distribution is sufficiently normal to use confidence intervals and hypothesis tests that are based on that approximation.
You can investigate the interplay between \(n\) and \(p\) and the shape of the sampling distribution by using simulations. Play around with the following app to investigate how the shape, center, and spread of the distribution of \(\hat{p}\) changes as \(n\) and \(p\) changes.
- Describe the sampling distribution of sample proportions at \(n = 300\) and \(p = 0.1\). Be sure to note the center, spread, and shape.
This distribution is similar to the graph. It appears symmetrical and has a center at .1. Taking a closer look though, we also see that it is bimodal, and has two peaks, with one at p= .08 and a count of about 430, and the other at about p= .105 and a count of about 690.
- Keep \(n\) constant and change \(p\). How does the shape, center, and spread of the sampling distribution vary as \(p\) changes. You might want to adjust min and max for the \(x\)-axis for a better view of the distribution.
Just as we learned throughout this lab, the shape changes based on a symmetrical, parabolic-like shape. At about 50% proportion, the peak is reached at a count of almost 600.
- Now also change \(n\). How does \(n\) appear to affect the distribution of \(\hat{p}\)?
Decreasing n leads to a wider spread for the distribution, but the shape remains relatively the same in that there is a peak at 50%, and the distribution is generally symmetrical. The larger the sample, the narrower the distribution becomes.
More Practice
For some of the exercises below, you will conduct inference comparing two proportions. In such cases, you have a response variable that is categorical, and an explanatory variable that is also categorical, and you are comparing the proportions of success of the response variable across the levels of the explanatory variable. This means that when using infer, you need to include both variables within specify.
- Is there convincing evidence that those who sleep 10+ hours per day are more likely to strength train every day of the week? As always, write out the hypotheses for any tests you conduct and outline the status of the conditions for inference. If you find a significant difference, also quantify this difference with a confidence interval.
H0:There is no significant association between high schoolers who sleep 10+ hours per dayand strength training everyday of the week. H1: High schoolers who sleep 10+ hours per day are more likely to strength train everyday of the week.
sleep_students <- yrbss %>%
filter(school_night_hours_sleep == "10+")
sleep_students %>%
mutate(exercises_7d = ifelse(physically_active_7d == 7, "yes", "no")) %>%
drop_na(exercises_7d) %>%
specify(response = exercises_7d, success = "yes") %>%
generate(reps = 1000, type = "bootstrap") %>%
calculate(stat = "prop") %>%
get_ci(level = 0.95)## # A tibble: 1 × 2
## lower_ci upper_ci
## <dbl> <dbl>
## 1 0.313 0.419ci_width_sleep<-0.4217252-0.3162939
print(ci_width_sleep)## [1] 0.1054313margin_of_error_sleep2 <- ci_width_sleep / 2 #confidence interval/2
round(margin_of_error_sleep2,2)## [1] 0.05With a 95% confidence interval between about .32-.42, we can say that between 32-42% of , or 37±5%, reported that they sleep at least 10 hours and go to the gym everyday. This is significant and is enough for us to accept the alternative hypothesis, and reject the null hypothesis.
- Let’s say there has been no difference in likeliness to strength train every day of the week for those who sleep 10+ hours. What is the probability that you could detect a change (at a significance level of 0.05) simply by chance? Hint: Review the definition of the Type 1 error.
A type 1 error occurs when a null hypothesis is rejected when it is actually true. There is a 5% probability of detecting a change by chance.
- Suppose you’re hired by the local government to estimate the proportion of residents that attend a religious service on a weekly basis. According to the guidelines, the estimate must have a margin of error no greater than 1% with 95% confidence. You have no idea what to expect for \(p\). How many people would you have to sample to ensure that you are within the guidelines?*Hint:* Refer to your plot of the relationship between \(p\) and margin of error. This question does not require using a dataset.
A sample of 1000 people. * * *